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EM - I


Unit – 1


Higher order derivatives and applications

Q 1. Compute limx→−2(3x2+5x9)

Solution:

First, use property 2 to divide the limit into three separate limits. Then use property 1 to bring the constants out of the first two. This gives,

Limx→−2(3x2+5x9) =limx→−2(3x2) + limx→−2(5x) limx→−2(9)

=3(2)2+5(2) (9)

= -7

 

Q 2. Find the limits of limx3 [x(x+2)].

Solution: limx3 [x(x+2)] = 3(3+2) = 3 x 5 = 15

 

Q 3: Differentiability Implies Continuity If f is differentiable at x0, then f is continuous at x0.

It turns out that equation (1) can be restated.

Definition. Derivatives in One Dimension - Alternate Definition

Solution: Let f : D R R and let x0 be an interior point of D.

Then f is differentiable at x0 if there is a number f (x0) such that (2) lim xx0 f(x) f(x0) (x x0)f (x0) |x x0| = 0 and the number f (x0) is called the derivative of f at x0.

We use this definition to create a definition of the derivative for functions of

Several variables. Definition. Derivatives in Higher Dimensions Let f : D R n R

And let P0 be an interior point of D. Then f is differentiable at P0 if there is a vector

f (P0) such that

lim PP0 f (P) f (P0) # » P0P  · f (P0) =0    

                               |P P0|

 

Q 4:Suppose there is a cRcR and a function E:RRE:RR, s.t. f(x)=f(a)+c(xa)+E(x)f(x)=f(a)+c(xa)+E(x) near aa and limxaE(x)xa=0limxaE(x)xa=0.

Then:

Limxaf(x)f(a)xac=limxaf(x)f(a)c(xa)xa=0limxaf(x)f(a)xac=limxaf(x)f(a)c(xa)xa=0

So that means: 

    f(x)f(x) exists and f(x)=cf(x)=c. Of course writing f(x)f(x) instead of cc before you know this is indeed begging the question.

 

Q 5:  For a function f: RR f: RR, suppose that there exist a real number AA such that for all xx in some neighbourhood of aa,

f(x)=f(a)+A(xa) +E(x)f(x)=f(a)+A(xa) + E(x)

And that

Limx aE(x)(xa) =0limxaE(x)(xa) =0

We say that ff is differentiable at aa. If ff is differentiable at aa we define f(a)f(a) by

f(a)=Af(a)=A

This definition is not circular and is useful since it can be generalized to the case where f: RnRm f: RnRm. In this case AA is a matrix instead of a real number (to be precise, AA is a linear operator

 

Q 6: Find the nth derivative of sin3 x

  Solution: we know that sin 3x= 3sin x 4sin3 x = sin3x=    

 

Differentiate   n    times w.r.t x,

(sin3 x) =   (3 sinx- sin3x)

                  =  (  -3n. Sin (3x+ nπ/2) +   3 sin (x+ nπ/2)) nϵz

 

Q 7: Find the nth derivative of sin 5x. Sin 3x.?

Solution: let y =   sin 5x.sin 3x= ( sin 5x.sin 3x)  

            y= (  cos 2x -cos8x)

           y= (cos 2x- cos8x)

Differentiate n times w.r.t x,

 Yn = (cos 2x - cos8x)

yn = ( 2 n (cos (2x+ nπ/2)- 8n.cos (8x + nπ/2)) nϵz.

Q 8: Successive n th derivative of nth elementary function ie., exponential

If y = ae n x + be –nx,then show that   y2= n2y

 Solution: Y= aenx + be-nx

  y 1 =  a.n.enx - b.n.e-nx

y2 = an2enx – bn2 e-nx = n2 (ae nx+ be –nx)

y2= n2y.

 

Q 9: If y= e-kx/2(a cosnx+ b sinnx) then show that., y2+ ky1+(n2+ k2/4) y =0

Solution: y= e-kx/2(a cosnx+ b sinnx)

Differentiating w.r.to. x.,

 

Y1 = e-kx/2(-an sin nx + bn cos nx) - k/2. y

Y1+ k/2. y = ne-kx/2 (-an sin nx + bn cos nx) (1)

Differentiating w.r.to x.,

Y2+ k/2. y1 = ne-kx/2 (-k/2) ( -an sin nx + bn cos nx) + n e-kx/2(-an  cosnx- bn  sinnx).

= -(k/2) (y1+ k/2 y)- n2 y = - (k/2 y1)- (k2/4) y- n2y.

  y2 + ky1 +(n2+ k2/4) y = 0.

 

Q 10: If Y 2) = log (x + 2)) then show that

(1 + x 2) y1 +x y =1

Solution: Y 2) = log (x + 2))

Differentiating w.r.to x.,

y.1/ 2  2 .2x+ 2. Y1.

 = (1+x) y + x y = . 1/x 2 . 2)  +x/ 2 =1

 

Q 11: If Y 2) = log (x + 2)) then show that

(1 + x 2) y1 + x y =1

Solution: Y 2) = log (x + 2))

Differentiating w.r.to x.,

y.1/ 2  2 .2x+ 2. Y1.

 = (1+x y +x y = . 1/x+ 2 . 2)  +x/ 2 =1

Q 12:Evaluate

Solution:

Let

By L – Hospital rule,

 

Q 13: Evaluate

Solution:

Let

By L – Hospital rule

 

Q 14: Evaluate

Solution:

Let

By L – Hospital rule

 

Q 15: Find the value of a, b if

Solution:

Let

By L – Hospital rule

… (1)

But

From equation (1)

 

Q 16: Evaluate

Solution:

Let

(By L – Hospital Rule)

 

Q 17: Evaluate

Solution:

Let

… 0o form

Taking log on both sides we get,

By L – Hospital Rule

i.e.

 

Q 18: Evaluate

Solution:

Let

Taking log on both sides,

By L – Hospital rule,

i.e.

 

Q 19: Evaluate

Solution:

Let

Taking log on both sides, we get

By L – Hospital Rule,